What is the integral of (5x-3)/(x(x-3)(x+1)) ?

The integral of (5x-3)/(x(x-3)(x+1)) is ln|x-3|-2ln|x+1|+ln|x|+C

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integral of (5x-3)/(x(x-3)(x+1))

\int \frac{5 x - 3}{x ( x - 3 ) ( x + 1 )} d x

ln ∣ x - 3 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ + C

Solution steps

\int \frac{5 x - 3}{x ( x - 3 ) ( x + 1 )} d x

Expand

\frac{5 x - 3}{x ( x - 3 ) ( x + 1 )} : \frac{5 x}{x ( x - 3 ) ( x + 1 )} - \frac{3}{x ( x - 3 ) ( x + 1 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5 x}{x ( x - 3 ) ( x + 1 )} d x - \int \frac{3}{x ( x - 3 ) ( x + 1 )} d x

\int \frac{5 x}{x ( x - 3 ) ( x + 1 )} d x = 5 (\frac{1}{4} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x + 1 ∣)

\int \frac{3}{x ( x - 3 ) ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣ - ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 3 ∣

= 5 (\frac{1}{4} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x + 1 ∣) - (\frac{3}{4} ln ∣ x + 1 ∣ - ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

Simplify

5 (\frac{1}{4} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x + 1 ∣) - (\frac{3}{4} ln ∣ x + 1 ∣ - ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : ln ∣ x - 3 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣

= ln ∣ x - 3 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣

Add a constant to the solution

= ln ∣ x - 3 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ + C

\int x cos (7 π x^{2}) d x

- (y + t) = \frac{d y}{d t}

\int 3^{5 x} d x

3 y^{^{''}} + 10 y^{^{'}} = 0

\frac{d y}{d x} = x^{6} y^{- 3}

What is the integral of (5x-3)/(x(x-3)(x+1)) ?
The integral of (5x-3)/(x(x-3)(x+1)) is ln|x-3|-2ln|x+1|+ln|x|+C